Given data
*The given distance is s = 8.13 km = 8130 m
*The given time is t = 33.5 s
*The given acceleration is a = 4.6 m/s^2
(1)
The formula for the speed at the first beginning of 33.5 seconds is given as
[tex]s=ut+\frac{1}{2}at^2[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} 8130=u(33.5)+\frac{1}{2}(4.6)(33.5)^2 \\ u=165.6\text{ m/s} \end{gathered}[/tex]Hence, the speed at the first beginning of 33.5 s is u = 165.6 m/s
(2)
The formula for the speed at the end of the 33.5 s is given as
[tex]v=u+at[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=(165.6)+(4.6)(33.5)_{} \\ =319.7\text{ m/s} \end{gathered}[/tex]Hence, the speed at the end of the 33.5 s is v = 319.7 m/s
A baseball is hit from theground straight up into the airwith a speed of 14.6 m/s. Howlong is the ball in air (time inseconds from ground-to-ground)?
Givens.
• Initial speed = 14.6 m/s.
,• FInal speed = 0 m/s (at highest point).
,• Gravity = 9.8 m/s^2.
First, find the time needed to reach the highest point.
[tex]\begin{gathered} v_f=v_0+gt \\ t=\frac{v_f-v_0}{g} \\ t=\frac{0-14.6\cdot\frac{m}{s}}{-9.8\cdot\frac{m}{s^2}} \\ t\approx1.49\sec \end{gathered}[/tex]It takes 1.49 seconds to reach the highest point.
The time that the baseball takes to reach the ground is double t because the trajectory is symmetrical, that is, it takes the same time to go from ground level to highest point than from highest point to ground level.
[tex]t_{\text{total}}=2\cdot1.49\sec =2.98\sec [/tex]Therefore, the baseball is 2.98 seconds in the air.
The mass number of an atom is found by1) adding the number of protons and neutrons.2) adding the number of protons and electrons.3) subtracting the number of protons from the number of neutrons.4) subtracting the number of protons from the number of electrons.
The mass number of an atom is the sum of number of protons present in the nucleus and number of neutrons present in the same nucleus.
[tex]M=\text{Protons}+\text{Neutrons}[/tex]Thus, 1st option is the correct answer.
How much heat must be removed from 750 grams of water at 0°C to form ice at 0°C?
ANSWER:
250500 J
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 750 grams
We can determine the amount of heat removed by taking into account the following image:
This means that to go from water to ice there is an absorbed radiation dose of 334 J/g, therefore, the heat removed is:
[tex]\begin{gathered} Q=m\cdot a \\ \\ \text{ We replacing:} \\ \\ Q=750\cdot334 \\ \\ Q=250500\text{ J} \end{gathered}[/tex]The heat removed is 250500 joules.
# 3. If a wire is connected to row 1, column a , and another to row 1, column c, are the two wires connected ?
In a breadboard, internal connections are made between groups of five holes in each row. That means that holes A, B, C, D, and E in row 1 are all connected, but for example, F and A are not connected.
Since 1A and 1C are in the same row and in the same five-hole group, they are connected.
Answer: The two wires are connected.
What is the relationship. Stern average kinetic energy of a gas and it’s temperature?
Average kinetic energy is directly proportional to its temperature-
E= f/2 N k T
E= energy
T= temperature
A human heart found to beat seventy five times in a minute. Calculate the beat frequency?
[tex]{ \green{ \tt{f = \frac{number \: of \: beats}{time \: taken}}}} [/tex]
[tex]{ \green{ \tt{number \: of \: beats = 75}}}[/tex]
[tex]{ \green{ \tt{time \: taken =1 \: min \: = 60 \: sec}}}[/tex]
[tex]{ \red{ \sf{f = \frac{ \cancel{75^{3}}}{ \cancel{ 60_{4} }}}}}[/tex]
[tex]{ \blue{ \boxed{ \purple{ \sf{f = \frac{3}{4} = 1.2 {s}^{ - 1}}}}}} [/tex]
___________________________________
[tex]{ \blue{ \sf{T = \frac{1}{f}}}} [/tex]
[tex]{ \blue{ \sf{T = \frac{1}{ \purple{ \sf{1.2}}}}}} [/tex]
[tex]{ \boxed{ \red{ \sf{T = 0.8 \: S}}}}[/tex]
In the graph of a bus his journey through town. Which of the following choices best represent where the bus is decelerating
Answer:
The bus is decelerating from B to C
Explanation:
The graph given shows the velocity of a bus at certain points in time. Higher the point on the graph, the higher the veocity. Now, the bus decelerates when its speed reduces from a higher value to a lower value. This we see as happening from point B to point C.
Hence, the bus decelerates from B to C.
At other points, you see the bus as either accelerating ( from O to A and D to E) or standing still (from A to B, C to D, and E to F). Only from B to C does the bus decelerate.
A ball player catches a ball 3.2 s after throwing it vertically upward.With what speed did he throw it? What height did it reach?Express your answers using two significant figures.
Since the time of flight of the ball is 3.2 seconds, that means for 1.6 seconds the ball was going upwards, then it reached the maximum height (where the velocity is zero), then went 1.6 seconds downwards, until it reaches the hand of the player again.
Using the formula below, we can find the initial velocity of the ball:
[tex]V=V_0+a\cdot t[/tex]Where V is the final velocity after t seconds, V0 is the initial velocity and 'a' is the acceleration.
Using V = 0, t = 1.6 s and a = -9.8 (gravity's acceleration), we have:
[tex]\begin{gathered} 0=V_0-9.8\cdot1.6\\ \\ V_0-15.68=0\\ \\ V_0=15.68\text{ m/s} \end{gathered}[/tex]Rounding the answer to two significant figures, we have an initial velocity of 16 m/s.
To find the maximum height, we can use the formula below:
[tex]\begin{gathered} \Delta S=V_0t+\frac{at^2}{2}\\ \\ \Delta S=15.68\cdot1.6-\frac{9.8\cdot1.6^2}{2}\\ \\ \Delta S=25.088-12.544\\ \\ \Delta S=12.544\text{ m} \end{gathered}[/tex]Rounding the answer to two significant figures, the maximum height is 13 m.
How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer: ___________ J (round to the nearest whole number)
According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:
[tex]W=\Delta K[/tex]Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:
[tex]K=\frac{1}{2}mv^2[/tex]Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:
[tex]\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\ \\ \therefore K\approx532,346J \end{gathered}[/tex]Therefore, the answer is: 532,346 J.
A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.5 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 6.30 N. 1. What is the density of the metal? 2. What did the cube weigh before you drilled the hole in it?
1.
First, let's calculate the volume of the cube:
[tex]V_1=6^3=216\text{ cm^^b3}[/tex]Now, let's calculate the volume of the cylinder drilled:
[tex]\begin{gathered} V_2=\frac{\pi d^2h}{4}\\ \\ V_2=\frac{3.14159\cdot2.5^2\cdot6}{4}\\ \\ V_2=29.45\text{ cm^^b3} \end{gathered}[/tex]So the volume of the cube after being drilled is:
[tex]\begin{gathered} V=V_1-V_2\\ \\ V=216-29.45\\ \\ V=186.55\text{ cm^^b3} \end{gathered}[/tex]If the weight is 6.3 N, let's find the mass:
[tex]\begin{gathered} W=m\cdot g\\ \\ 6.3=m\cdot9.8\\ \\ m=\frac{6.3}{9.8}\\ \\ m=0.643\text{ kg} \end{gathered}[/tex]And the density of the metal is:
[tex]\begin{gathered} d=\frac{m}{V}\\ \\ d=\frac{0.643}{186.55}\\ \\ d=0.0034468\text{ kg/cm^^b3}\\ \\ d=3.4468\text{ g/cm^^b3} \end{gathered}[/tex]2.
To find the weight of the cube before being drilled, let's use the following rule of three:
[tex]\begin{gathered} volume\rightarrow weight\\ \\ 216\text{ cm^^b3}\rightarrow x\text{ N}\\ \\ 186.55\text{ cm^^b3}\rightarrow6.3\text{ N}\\ \\ \\ \\ \frac{216}{186.55}=\frac{x}{6.3}\\ \\ x=\frac{216\cdot6.3}{186.55}\\ \\ x=7.295\text{ N} \end{gathered}[/tex]A person pulls a block 3 m along a horizontal surface by a constant force F=20 N. Determine the work done by force F acting on the block.
Take into account that the work done is given by the following formula:
[tex]W=F\cdot d[/tex]where,
F: applied force = 20N
d: distance = 3m
Replace the previous values into the formula for W:
[tex]W=(20N)(3m)=60J[/tex]Hence, the work done by the person is 60J
A 0.2-kg aluminum plate, initially at 20°C, slides down a 15-m-long surface, inclined at a 30°angle to the horizontal. The force of kinetic friction exactly balances the component ofgravity down the plane so that the plate, once started, glides down at constant velocity. If90% of the mechanical energy of the system is absorbed by the aluminum, what is itstemperature increase at the bottom of the incline? (Specific heat for aluminum is 900J/kg⋅°C.) Why do I multiply 15 by sin30?
A scheme of a the given situation is shown below:
First, consider that the work over the plate is done only by the component of the weight parallel to the incline (due to the perpendicular component is balanced by the friction force), then, the work on the plate is:
W = m*g*d*sinθ
where,
m: mass = 0.2kg
d: length of the incline = 15m
g: gravitational acceleration constant = 9.8m/s^2
θ = 30
By replacing the previous values into the expression for W, you obtain:
W = (0.2 kg)(9.8 m/s^2)(15 m)sin(30)
W = 14.7 J
Now, take into account that the amount of heat absorbed by the aluminum plate is given by the following formula:
Q = m*c*ΔT
Q: heat
m: mass
c: specific heat
ΔT: change in tempetaure
Take into account that the 90% of the mechanical energy is absorbed by the plate, which means that 0.9 of the work is converted to absorbed heat by the plate.
Then, you can write:
0.9W = Q
0.9(14.7J) = Q
13.23J = Q
Replace the given expression for Q into the previous equation and solve for ΔT, as follow:
m*c*ΔT = 13.23 J
ΔT = 13.23J/(m*c)
Now, replace the values of m and c for aluminum and simplify:
ΔT = 13.23J/(0.2kg*900J/kg°C)
ΔT = 0.0735°C
Hence, the temperature increase at the bottom of the incline is approximately 0.07°C
Magnetic field lines of force will not go through ______A) your hand B) a tree C) a steel plate D) a paper plate E) water
Magnetic field lines can pass through metals and non metals. But it will not pass through super conductors. A steel plate is a superconductor. Thus, the correct option is
C) a steel plate
Part 1)If a force of magnitude 125 N is exerted horizontally as shown, find the force exerted by the hammer claws on the nail. (Assume that the force the hammer exerts on the nail is parallel to the nail).Answer in units of N.Part 2)Find the force exerted by the surface on the point of contact with the hammer head. Assume that the force the hammer exerts on the nail is parallel to the nail. Answer in units of N.
Given data:
* The force acting on the hammer in the horizontal direction is F = 125 N.
* The distance of horizontal force from the point of contact is d = 28 cm.
* The angle of the nail with the vertical direction is,
[tex]\theta=26^{^{\circ}}[/tex]* The distance of the nail from the point of contact is x = 5.86 cm.
Solution:
The moment caused by the horizontal force on the hammer is equal to the moment caused by the horizontal component of force at the nail about the point of contact, thus,
[tex]F\times d=F^{\prime}x\text{ sin(}\theta)[/tex]Substituting the known values,
[tex]\begin{gathered} 125\times28=F^{\prime}\times5.86\times\sin (26^{\circ}) \\ 3500=F^{\prime}\times2.57 \\ F^{\prime}=\frac{3500}{2.57} \\ F^{\prime}=1361.87\text{ N} \end{gathered}[/tex]Thus, the force exerted on the nail is 1361.87 N or approximately 1362 N.
A bowling ball of mass 7.29 kg and radius 11.0 cm rolls without slipping down a lane at 3.00 m/s. Calculate the total kinetic energy.
We have the next information
m=7.29 kg
v=3 m/s
r=11cm=0.11m
We can find the kinetic energy using the next formula
[tex]KE=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]I is the moment of inertia
w is the angular velocity
First, we need to calculate the moment of inertia
[tex]I=\frac{2}{5}mr^2[/tex][tex]I=\frac{2}{5}(7.29)(0.11)^2=0.0353kgm^2[/tex]Then for the angular velocity
[tex]\omega=\frac{v}{r}[/tex][tex]\omega=\frac{3}{0.11}=27.28[/tex]then we will substitute the values in the kinetic energy formula
[tex]KE=\frac{1}{2}(7.29)(3)^2+\frac{1}{2}(0.0353)(27.28)^2[/tex][tex]KE=45.9J[/tex]The total kinetic energy is 45.9 J
i hve attached the question
The speed of the planet Venus 3.56 × 10⁵ m/s.
We are given that,
The radius of planet Venus = r = 1.08 × 10¹¹ m
The orbital time period = t = 225 days = 1.9 × 10⁶sec
The speed of planet Venus = v = ?
The formula of the time period of complete one revolution of planet to the sun is given as,
T = 2π/ω
ω = r.v
v = 2πr/T
Putting , the values in above equation we can get,
v = (2 × 3.14 × 1.08 × 10¹¹ m)/ ( 1.9 × 10⁶sec)
v = 3.56 × 10⁵ m/s
Therefore , the speed of planet Venus would be 3.56 × 10⁵ m/s
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17. What distinguishes single, double, and triple
covalent bonds?
Answer:
The number of shared electrons is the major distinction between single double and triple bonds. A single bond is formed when two atoms share one pair of electrons, whereas a double bond is formed when two atoms share two pairs (four electrons). Three pairs of electrons (six atoms) are shared to form triple bonds.
Answer
In a single bond one pair of electrons is shared, with one electron being contributed from each of the atoms. Double bonds share two pairs of electrons and triple bonds share three pairs of electrons. Bonds sharing more than one pair of electrons are called multiple covalent bonds.
Explanation:
I need help with this table pleasecalculate relative density of steel. Use table 3
Take into account that the relative density is given by:
[tex]\rho_{\text{rel}}=\frac{\rho}{\rho_{\text{water}}}[/tex]where ρ, in this case, is the density of the steel and ρwater is the density of water (1000 kg/m^3).
The density of the steel is:
[tex]\rho=\frac{\text{mass}}{\text{volume}}[/tex]Based on table 3, you have:
mass = 50.7 g = 0.0507 kg
volume = 0.0000063 m^3
[tex]\rho=\frac{0.0507kg}{0.0000063m^3}\approx8047.62\frac{kg}{m^3}[/tex]Then, for the relative density you obtain:
[tex]\rho_{\text{rel}}=\frac{\rho}{\rho_{\text{water}}}=\frac{8047.62\frac{kg}{m^3}}{1000\frac{kg}{m^3}}\approx8.048[/tex]Hence, the relative density of steel is 8.048
This question is based on Oscillations and waves. I tried it for days and I just couldn't get it right.
ANSWER:
The maximun velocity is 16.07 m/s
At x = 0.26
The velocity is 8.36 m/s
The accelearion is 286.67 m/s^2
The resorting force is 86 N
STEP-BY-STEP EXPLANATION:
Given:
k = 310 N / m
Max distance = 0.5 m
Mass of block = 0.3 kg
Max velocity:
Using conservation of energy:
[tex]\begin{gathered} \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v^2=\frac{kx^2}{m} \\ \text{ replacing} \\ v^2=\frac{310\cdot0.5^2}{0.3} \\ v=\sqrt[]{258.33} \\ v=16.07\text{ m/s} \end{gathered}[/tex]At x = 0.26 m:
[tex]\begin{gathered} v^2=\frac{kx^2}{m} \\ v^2=\frac{310\cdot0.26^2}{0.3} \\ v=\sqrt[]{69.85} \\ v=8.36\text{ m/s} \end{gathered}[/tex]Acceleration:
[tex]\begin{gathered} F=k\cdot x \\ F=m\cdot a \\ \text{ therefore} \\ m\cdot a=k\cdot x \\ a=\frac{k\cdot x}{m} \\ \text{ replacing} \\ a=\frac{310\cdot0.26}{0.3} \\ a=286.67\text{ }\frac{m}{s^2} \end{gathered}[/tex]The resorting force:
[tex]\begin{gathered} F=m\cdot a \\ \text{ replacing} \\ F=0.3\cdot286.67 \\ F=86\text{ N} \end{gathered}[/tex]1 pts
An coconut with a mass of 2 kg and a feather with a mass of 0.01 kg fall from a tree through the air to the ground
below, both eventually reaching terminal velocity. At terminal velocity, the amount of air-rresistance force what is the answer ? A grather on the coconut B the same on each C grather on the feather
Take into account that air-resistance force is greater against bodies with lower densities.
In this case, the feather has a lower density than the coconut, then, you can conclude that air-resistance force is greater on the feather (option C).
Can anyone help me in this question, Please?
The net force acting on the mass [tex]m_{1}[/tex] is 12 N and the tension (T) in the string is 42 N.
Both the blocks will move with the same acceleration, that is, a = 3 [tex]m/s^{2}[/tex]
Now, from the free-body diagram of the block [tex]m_{1}[/tex]
T - F = [tex]m_{1} a[/tex]
It is given that F = 30 N
and [tex]m_{1} = 4 kg[/tex] , a = 3[tex]m/s^{2}[/tex]
Putting all these values, we get:
T - 30 = 4*3 = 12
T = 12 + 30
T = 42 N
Hence, the tension in the string = 42 N
Now, the net force acting on the mass [tex]m_{1}[/tex] is
[tex]F_{net}[/tex] = T - F = 42 - 30 = 12 N
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A truck can travel at 100 km/hr How long would it take to drive 900km?
Given:
The speed of the truck is,
[tex]v=100\text{ km/hr}[/tex]The distance is,
[tex]s=900\text{ km}[/tex]The time to drive this distance is,
[tex]t=\frac{s}{d}[/tex]Substituting the values we get,
[tex]\begin{gathered} t=\frac{900}{100} \\ =9\text{ hrs} \end{gathered}[/tex]Hence, the time is 9 hrs.
The rumble feature on a video game controller is driven by a device that turns electrical energy into mechanical energy. This device is best referred to as _________?A. an electric generatorB. an electromagnetC. a solenoidD. a motor
The rumble feature on a video game controller is driven that turns electrical energy into mechanical energy. This device is best referred to as a motor.
An electric motor converts electrical energy into mechanical energy.
Thus, the option (D) is correct.
Water flows through a pipe diameter of 8.000 cm at 49.0 m/min. Find the flow rate in m3/min
We are asked to determine the volumetric flow rate through a pipe of diameter 8.000 cm. To do that we will use the following formula:
[tex]R=Av[/tex]Where:
[tex]\begin{gathered} R=\text{ volumetric flow rate} \\ A=\text{ cross-area of the pipe} \\ v=\text{ velocity of the flow} \end{gathered}[/tex]The cross-area of the pipe is the area of a circle and is given by:
[tex]A=\frac{\pi D^2}{4}[/tex]Where:
[tex]\begin{gathered} A=\text{ cross-area} \\ D=\text{ diameter} \end{gathered}[/tex]Before we determine the area we will convert the diameter from cm to meters using the following conversion factor:
[tex]100cm=1m[/tex]Multiplying by the conversion factor we get:
[tex]8.000cm\times\frac{1m}{100cm}=0.080m[/tex]Now, we plug in the value in the formula for the area:
[tex]A=\frac{\pi(0.080m)^2}{4}[/tex]Solving the operations:
[tex]A=0.005m^2[/tex]Now, we plug in the values of area and velocity in the formula or the volumetric flow rate:
[tex]R=(0.005m^2)(49.0\frac{m}{\min })[/tex]Solving the operations:
[tex]R=0.246\frac{m^3}{min}[/tex]Therefore, the flow rate is 0.246 cubic meters per minute.
3. Free Fall: A ball is dropped from a window that is 35 meters above gound level.How fast is the ball traveling when it reaches the ground?How long does the ball take to reach the ground?
We are given that a ball is dropped from a height of 35 meters in a free fall and we are asked to determine its final velocity. To do this we will use the following equation of motion:
[tex]v^2_f-v^2_0=2ah[/tex]Where:
[tex]\begin{gathered} v_f\text{ = final velocity} \\ v_0=\text{ initial velocity} \\ a=\text{ acceleration} \\ h=\text{ height} \end{gathered}[/tex]Since the ball is dropped this means that the initial velocity is zero. Also, since we have a free-fall motion the acceleration is the same as the acceleration of gravity. Replacing these values we get:
[tex]v^2_f=2gh[/tex]Now we solve for the velocity by taking the square root to both sides:
[tex]v_f=\sqrt[]{2gh}[/tex]Now we replace the values for the gravity and height:
[tex]v_f=\sqrt[]{2(9.8\frac{m}{s^2})(35m)}[/tex]Solving the operations:
[tex]v_f=26.19\frac{m}{s}[/tex]Therefore, the final velocity is 26.19 meters per second.
Now we will determine the time it takes the ball to reach the ground. To do that we will use the following equation of motion:
[tex]v_f=v_0+gt[/tex]Since the initial velocity is zero the equation simplifies to:
[tex]v_f=gt[/tex]Now we divide by the acceleration of gravity:
[tex]\frac{v_f}{g}=t[/tex]Now we replace the values:
[tex]\frac{26.19\frac{m}{s}}{9.8\frac{m}{s^2}}=t[/tex]Solving the operations we get:
[tex]2.67s=t[/tex]Therefore, it takes 2.67s for the ball to reach the ground in free-fall.
Help me with number 1 I’m very lost. I just need the equation. No explanation I’m stuck on 1 part.
Given data:
* The value of angular velocity is,
[tex]\omega=1.3\text{ rad/s}[/tex]Solution:
(a). The time period of the oscillation in terms of the angular velocity is,
[tex]T=\frac{2\pi}{\omega}[/tex]Substituting the known values,
[tex]\begin{gathered} T=\frac{2\pi}{1.3} \\ T=4.83\text{ s} \end{gathered}[/tex]Thus, the time period of oscillation is 4.83 s.
(b). The frequency of the oscillation in terms of the time period is,
[tex]undefined[/tex]As a sports psychologist how would you suggest working with an athlete who suffers from self-
attention in front of the home crowd?
As a sports psychologist my suggestion would be to ensure that they maintain a balanced approach about others opinions and should concentrate on their performance than the opinion of other.
Many athletes are intimidated by or motivated by the support and encouragement of the home audience when competing in front of them. They might make an effort to get people's respect and favor. Although it can improve performance, it can also negatively increase an athlete's pressure to perform or even cause them to feel overconfident. Athletes could feel humiliated if they lose or fail because they couldn't satisfy the expectations of their audience. Additionally, they can lose self-respect, which might have an effect on their morale and future performance.
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What is the resistance of an electric frying pan that draws 12 amperes of current when connected to 120 Volt circuit
10 ohms
Explanation
the resistance of an electric circuit is given by:
[tex]\begin{gathered} R=\frac{V}{I} \\ where\text{ R is the resitance} \\ Vi\text{s the voltage} \\ I\text{ is the current} \end{gathered}[/tex]so
Step 1
a) let
[tex]\begin{gathered} R=R \\ V=120\text{ volts} \\ I=12\text{ Amperes} \end{gathered}[/tex]b) now, replace in the expression
[tex]\begin{gathered} R=\frac{V}{I} \\ R=\frac{120\text{ volt}}{12\text{ Amp}} \\ R=10\text{ ohms} \end{gathered}[/tex]therefore, the answer is
10 ohms
I hope this helps you
15) A closed, uninsulated system was fitted with a movable piston. The addition of 754.3 J of heat caused the system to expand, doing 424.5 J of work in the process against a constant pressure. What is the value of E for this process?
Given,
Addition of heat, Q=+754.3 J
Work done , W=-424.5 J
The work is done against constant pressure.
To find
The value of E for this process
Explanation
According to the first law of thermodynamics,
[tex]\Delta E=Q+W[/tex]Putting the values,
[tex]\begin{gathered} \Delta E=754.3-424.5 \\ \Rightarrow\Delta E=329.8J \end{gathered}[/tex]Conclusion
The required value is 329.8 J
What does the strength of frictiondepend on?A. The direction of the forces.B. The types of surfaces and how hard theobjects are being pushed.C. The color of surfaces and how hard they push.D. Only how hard the objects are being pushed.
Answer:
B. The types of surfaces and how hard the objects are being pushed.
Explanation:
The force of friction is the result of the